hdu 5178 pairs【尺取】

John has nn points on the X axis, and their coordinates are (x[i],0),(i=0,1,2,…,n−1)(x[i],0),(i=0,1,2,…,n−1). He wants to know how many pairs<a,b><a,b> that |x[b]−x[a]|≤k.(a<b)|x[b]−x[a]|≤k.(a<b)

Input

The first line contains a single integer TT (about 5), indicating the number of cases. 
Each test case begins with two integers n,k(1≤n≤100000,1≤k≤109)n,k(1≤n≤100000,1≤k≤109). 
Next nn lines contain an integer x[i](−109≤x[i]≤109)x[i](−109≤x[i]≤109), means the X coordinates.

Output

For each case, output an integer means how many pairs<a,b><a,b> that |x[b]−x[a]|≤k|x[b]−x[a]|≤k.

Sample Input

2
5 5
-100
0
100
101
102
5 300
-100
0
100
101
102

Sample Output

3
10

注意ll型并且输入要用scanf，cin的话卡着时间过的，scanf快了好多

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[100005];
int main()
{
	ll  t;	
	ll i,j;
	scanf("%lld",&t);
	while(t--)
	{
		ll n,k;
		scanf("%lld%lld",&n,&k);
		for(i=0;i<n;i++){
			scanf("%lld",&a[i]);
		}
		sort(a,a+n);
		ll ans=0;
		for(i=0,j=0;i<n;i++){
			while(a[j+1]-a[i]<=k && j+1<n)//每次从当前的i位置开始向后找
				j++;
			ans+=j-i;
		}
       // 	for(i=0;i<n;i++){
		//	int pos=i;
		//	for(j=pos;a[j+1]-a[i]<=k && j<n-1;j++)
		//	{}
		//	ans+=j-i;
		//}  不知道为什么换成for之后一直超时。。。
		cout<<ans<<endl;
	}
	return 0;
}